Let $||.||_{\alpha}$ be a C*-norm on $A\odot B$, $A\otimes_{\alpha} B$ be the completion and $\xi$ be a state on $A\otimes_{\alpha} B$. We definte the restrictions $\xi|_{A}$ and $\xi|_{B}$ as follows. Let ($\pi_{\xi}, H_{\xi}, v_{\xi}$) be the GNS triplet and $\pi_{\xi, A}$ and $\pi_{\xi, B}$ be the restriction homomorphisms on $A$ and $B$ respectively. Then define $\xi|_{A}(a)=\langle\pi_{\xi, A}(a)v_{\xi}, v_{\xi}\rangle$ and $\xi|_{B}(a)=\langle\pi_{\xi, B}(b)v_{\xi}, v_{\xi}\rangle$.
If we have $\pi_{\xi, A}(A)\subset \mathbb{C}1$ (here $1$ denotes the identity in $B(H_{\xi})$), then can we verify that $\xi|_{A}$ is a pure state?
Yes. Because $\pi_{\xi,A}$ is a representation, thus multiplicative. As $v_\xi$ has norm one, $$ \xi|_A(a_1a_2)=\langle \pi_{\xi,A}(a_1a_2)v_\xi,v_\xi\rangle=\langle \pi_{\xi,A}(a_1)\pi_{\xi,A}(a_2)v_\xi,v_\xi\rangle=\langle \pi_{\xi,A}(a_1)v_\xi,v_\xi\rangle\,\langle \pi_{\xi,A}(a_2)v_\xi,v_\xi\rangle=\xi|_A(a_1)\xi|_A(a_2). $$ And any multiplicative state is pure. To see this let $\varphi$ be a multiplicative state of the C$^*$-algebra $A$. Let $(\pi,H,v)$ be the GNS representation of $\varphi$. Then, for any $a\in A$, \begin{align*} \|\pi(a)v-\varphi(a)v\|^2&=\|\pi(a)\|^2+|\varphi(a)|^2-2\text{Re}\,\langle\pi(a)v,\varphi(a)v\rangle\\ &=\langle\pi(a^*a)v,v\rangle+|\varphi(a)|^2-2\text{Re}\,\overline{\varphi(a)}\,\langle \pi(a)v,v\rangle\\ &=\varphi(a^*a)+|\varphi(a)|^2-2\text{Re}\,\overline{\varphi(a)}\varphi(a)\\ &=2|\varphi(a)|^2-2|\varphi(a)|^2\\ &=0. \end{align*} So $\pi(a)v=\varphi(a)v$ for all $a\in A$, showing that $H=\mathbb C\,v$, one-dimensional. Now we can deduce that $\pi$ is irreducible, implying that $\varphi$ is pure.